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July 15, 2023
Question 1. Find the amount if Rs 50000 is invested at 10% p.a. for 4 years
Options:
A. 73205
B. 7320.5
C. 73250
D. 73502
Solution: We know that,
Amount = P (1+ \frac{r}{100})^{n}
A =50,000(1+ \frac{10}{100})^{4}
A = 50000 (1.1)4
A = 73205
Correct option: A
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Question 1. The Compound Interest on a sum of Rs 576 is Rs 100 in two years. Find the rate of interest.
A. 7.33%B. 4.33%C. 8.33%D. 5.33%
Amount = Compound Interest + Principal
A = 576 + 100 = 676
Amount =P (1+ \frac{r}{100})^{n}
676 =576 (1+ \frac{r}{100})^{2}
\frac{676}{576}=(1+ \frac{r}{100})^{2}
(\frac{26}{24})2 = (1+ \frac{r}{100})^{2}
\frac{26}{24}=(1+ \frac{r}{100})
(\frac{26}{24} – 1) =\frac{r}{100}
1.0833 – 1 =\frac{r}{100}
0.0833 = \frac{r}{100}
r = 8.33%
Correct option: C
Question 1. The difference between the CI and SI on a certain amount at 14% p.a. for 2 years is Rs. 100. What will be the value of the amount at the end of three years if compounded annually?
A. 7558.89B. 7500.45C. 7558D. 7554.56
Solution: We know that, Difference between CI and SI for 2 years , Difference = \frac{P(R)^{2}}{(100)^{2}}
100 = \frac{P(14)^{2}}{(100)^{2}}
P =\frac{100(100)^{2}}{(14)^{2}}
P =\frac{1000000}{196}
P = 5102.04
Now calculate the CI on Rs. 5102.04
A = 5102.04 (1+ \frac{14}{100})^{3}
A =5102.04 ( \frac{114}{100})^{3}
A = 5102.04 * 1.48
A= 7558.89
Question 1. A sum of money placed at compound interest doubles itself in 8 years. In how many years will it amount to 32 times itself?
A. 40 yearsB. 50 yearsC. 32 yearsD. 35 years
x^{\frac{1}{a}} = y^{\frac{1}{b}}
2^{\frac{1}{8}} = 32^{\frac{1}{x}}
2^{\frac{1}{8}} = 2^{\frac{5}{x}}
\frac{1}{8} = \frac{5}{x}
x = 40 years
Question 1. Find the compound interest on a sum of money of Rs.10000 after 2 years, if the rate of interest is 2% for the first year and 4% for the next year?
A. Rs. 304B. Rs. 608C. Rs. 1000D. Rs. 710
Amount = P (1+\frac{r_{1}}{100}) (1+\frac{r_{2}}{100}) (1+\frac{r_{3}}{100})
Here, r1 = 2% r2 = 4% and p = Rs.10000,
CI = A – P
CI = 10000 (1 + \frac{2}{100}) (1 + \frac{4}{100}) – 10000
CI = 10000 * (\frac{102}{100})(\frac{104}{100}) – 10000
CI = 10000 * (51 * \frac{52}{2500}) – 10000
CI = 10000 * (\frac{2652}{2500}) – 10000
CI = 10000 * (1.06) – 10000
CI = 10608 – 10000
CI = 608
Hence, the required compound interest is Rs. 608.
Correct option: B
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